An electron has been accelerated by a potential difference of 100 V.If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty ($\Delta$p/p *100) of the electron's momentum?
A)1%
B)2%
C)10%
D)10%
E)5%

Question

An electron has been accelerated by a potential difference of 100 V.If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty ($\Delta$p/p *100) of the electron's momentum?
A)1%
B)2%
C)10%
D)>>10%
E)5%

Quizplus · Accepted Answer

The de Broglie wavelength of the electron is given by $\lambda = h/p$, where h is Planck's constant and p is the momentum of the electron. The uncertainty in the electron's position is $\Delta x = 1 nm$, so the uncertainty in its momentum is $\Delta p = h/\Delta x$. The percent uncertainty in the electron's momentum is therefore $\Delta p/p *100 = (h/\Delta x)/p *100 = (h/p)/(\Delta x/p) *100 = (h/p)/(\lambda) *100$. Since $\lambda$ is much smaller than 1 nm, the percent uncertainty in the electron's momentum is much greater than 10%.

An Electron Has Been Accelerated by a Potential Difference of 100